I could use a hand to hold, i am a drift with this concept.

find a general solution to this functional differential equation.

f ''(x) = f(f(x)) and there are general solutions

In other words find a function whose 2nd derivative is equal to the same function evaluated at all the points that make that function up.

For example f(f(x)) = ? when f(x) = x^2
is for x=2 ----> f(f(4)) ----> f(4) = 16 ----> f(16) = 256 and this has to be equal to its 2nd derivative
y=0 meets these criteria, I think.
Let's say your function is y = Ax^2 + Bx + C

I think you're looking for a solution where

2A = A(Ax^2+Bx+C)^2 + B(Ax^2+Bx+C) + C
y=e^x is probably the answer they are looking for
so y'=2ax+b
i guess b in this instance would be my constant so i would drop the c
>>4
I am not sure, after the first der I would include a constant to the function
y = Ax^3

So you're looking for

6Ax = A(Ax^3)^3

6x = (Ax^3)^3

6x = (Ax^3)(Ax^3)(Ax^3)

(6/A)x = x^9

6/A = x^8

A = 6/(x^8)

Plug it in the original equation

y = 6x^3 / x^8

y = 6 / x^5
>>7
thank you, for trying I must be missing a building block
I guess the constant gets dropped.
>>7
f ''(x) = f(f(x))
I think this was my hang up, neat.
>>6
f(x) = e^x is the only function that is itself primed, if i remember right

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